# How to Find Pi Using Regular Polygons

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I am a former maths teacher and owner of DoingMaths. I love writing about maths, its applications, and fun mathematical facts.

## What Is Pi?

If you take any perfect circle and measure its circumference (the distance around the edge of the circle) and its diameter (the distance from one side of the circle to the other, going through the centre) and then divide the circumference by the diameter, you should find that you get an answer of approximately 3.

If you could make your measurements perfectly accurate, you would find that you actually get an answer of 3.14159... regardless of what size your circle is. It wouldn't matter if you were taking your measurements from a coin, the centre circle of a football pitch or even from the O2 Arena in London; as long as your measurements are accurate, you will get the same answer: 3.14159...

We call this number 'pi' (denoted by the Greek letter π), and it is sometimes also known as the Archimedes constant (after the Greek mathematician who first tried to calculate the exact value of pi).

Pi is an irrational number which mathematically means that it cannot be written as a fraction of two whole numbers. This also means that the digits of pi are never ending and never repeat themselves.

Pi has many applications for mathematicians, not just in geometry but throughout many other areas of mathematics as well. Its link to circles is also a valuable tool in many other areas of life, including science, engineering, etc.

In this article, we will look at a simple geometrical way of calculating pi using regular polygons.

## Unit Circle

Consider a unit circle such as in the picture above. Unit means that it has a radius equal to one unit (for our purposes, it doesn't matter what this unit is. It could be m, cm, inches, etc. The result will still be the same).

The area of a circle is equal to π x radius2. As the radius of our circle is one, we, therefore, have a circle with an area of π. If we can then find the area of this circle using a different method, we have therefore got ourselves a value for π.

## Adding Squares to our Unit Circle

Now imagine adding two squares to our picture of the unit circle. We have a larger square, just big enough for the circle to fit inside perfectly, touching the square in the centre of each of its edges.

We also have a smaller, inscribed square which fits inside the circle and is just big enough that its four corners all touch the edge of the circle.

It is clear from the picture that the area of the circle is smaller than that of the big square but larger than that of the small square. Therefore if we can find the squares' areas, we will have upper and lower bounds for π.

The large square is relatively simple. We can see that it is twice the width of the circle, so each edge is 2 long. The area is therefore 2 x 2 = 4.

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The smaller square is a little trickier as this square has a diagonal of 2 instead of an edge. Using Pythagoras theorem if we take a right-angled triangle made of two of the square's edges and the diagonal as the hypotenuse, we can see that 22 = x2 + x2 where x is the length of one edge of the square. This can be solved to get x = √2, hence the area of the small square is 2.

As the area of the circle is in between our two area values, we now know that 2 < π < 4.

## Unit Circle with Pentagons

So far, our estimate using squares isn't very precise, so let's see what happens if we start using regular pentagons instead. Again, I've used a larger pentagon on the outside with the circle just touching its edges, and a smaller pentagon on the inside with its corners just touching the edge of the circle.

Finding the area of a pentagon is a little trickier than for a square but not too difficult using trigonometry.

## Area of the Larger Pentagon

Take a look at the diagram above. We can split the pentagon up into ten equal right-angled triangles, each having a height of 1 (the same as the radius of the circle) and a centre angle of 360 ÷ 10 = 36°. I have denoted the edge opposite the angle as x.

Using basic trigonometry, we can see that tan 36 = x/1, so x = tan 36. Therefore, the area of each of these triangles is 1/2 x 1 x tan 36 = 0.3633. As there are ten of these triangles, the pentagon's area is 10 x 0.363 = 36.33.

## The Area of the Smaller Pentagon

The smaller pentagon has a distance of one from the centre to each vertex. We can split the pentagon up into five isosceles triangles, each with two edges of 1 and an angle of 360 ÷ 5 = 72°. The area of the triangle is therefore 1/2 x 1 x 1 x sin 72 = 0.4755, giving us a pentagon area of 5 x 0.4755 = 2.378.

We now have more accurate bounds for π of 2.378 < π < 3.633.

## Using Regular Polygons With More Sides

Our calculation using the pentagons is still not very precise, but it can be seen clearly that the more sides the polygons have, the closer together the bounds become.

We can generalise the method we used to find the pentagon areas to enable us to quickly calculate the inner and outer polygons for any number of sides.

Using the same method as for the pentagons, we get:

Area of smaller polygon = 1/2 x n x sin (360/n)

Area of larger polygon = n x tan (360/2n)

where n is the number of sides of the polygon.

We can now use this to get much more precise results! Upper and lower bounds using polygons with more sides

## Polygons With More Sides

Above I have listed the results for the next five polygons. You can see that the bounds get closer and closer together each time until we have a range of slightly over 0.3 when using decagons. This still isn't overly precise though. How many edges will we need to have before we can calculate π to 1 d.p and beyond?